Surface integral of position vector over a sphere












0












$begingroup$


$iint_S$ r.n $dS$



Over the surface of the sphere with radius $a$ centered at the origin



Now this is obviously trivial and the answer is $4pi a^3$ but I want to do it the hard way because there's something I don't understand



The surface is $x^2 + y^2 + z^2 = a^2$ , then the normal vector $n = nabla S$



$hat n$ = $frac{nabla S}{|nabla S|}$ = $frac{x hat i + y hat j + z hat k}{a}$



$dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$



Then $iint_S$ r.n $dS$ = $iint_S frac{x^2 + y^2}{sqrt{a^2 -x^2 -y^2}} + sqrt{a^2 -y^2 -x^2}$ $dxdy$



Switching to polar coordinates, $x=rho cosphi , y =rho sinphi$



Then $iint_S$ r.n $dS$ = $iint_S frac{rho^2}{sqrt{a^2 -rho^2}} + sqrt{a^2 - rho^2}$ $rho drho dphi$



Integrating $rho$ from $0$ to $a$ and $phi$ from $0$ to $2pi$ , we get:



$iint_S$ r.n $dS$ = $2pi a^3$ which is half the required answer $4pi a^3$ , is it because I only took into account that $dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$ and should have changed this surface element starting from a specific point? If so, how? Thanks






vector-analysis surface-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    When you take the square root, aren’t you only giving the correct value of $z$ in upper half space?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:10










  • $begingroup$
    Yes, but didn't I integrate from 0 to $2pi$ anyway?
    $endgroup$
    – khaled014z
    Dec 25 '18 at 2:25






  • 1




    $begingroup$
    This seems to be full of errors. Why are you varying $rho$. It's constant, you're on a sphere. The surface element should be $a^2sinthetaoperatorname dthetaoperatorname dvarphi $. You need spherical coordinates, not polar. Etc...
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 3:07
















0












$begingroup$


$iint_S$ r.n $dS$



Over the surface of the sphere with radius $a$ centered at the origin



Now this is obviously trivial and the answer is $4pi a^3$ but I want to do it the hard way because there's something I don't understand



The surface is $x^2 + y^2 + z^2 = a^2$ , then the normal vector $n = nabla S$



$hat n$ = $frac{nabla S}{|nabla S|}$ = $frac{x hat i + y hat j + z hat k}{a}$



$dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$



Then $iint_S$ r.n $dS$ = $iint_S frac{x^2 + y^2}{sqrt{a^2 -x^2 -y^2}} + sqrt{a^2 -y^2 -x^2}$ $dxdy$



Switching to polar coordinates, $x=rho cosphi , y =rho sinphi$



Then $iint_S$ r.n $dS$ = $iint_S frac{rho^2}{sqrt{a^2 -rho^2}} + sqrt{a^2 - rho^2}$ $rho drho dphi$



Integrating $rho$ from $0$ to $a$ and $phi$ from $0$ to $2pi$ , we get:



$iint_S$ r.n $dS$ = $2pi a^3$ which is half the required answer $4pi a^3$ , is it because I only took into account that $dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$ and should have changed this surface element starting from a specific point? If so, how? Thanks






vector-analysis surface-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    When you take the square root, aren’t you only giving the correct value of $z$ in upper half space?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:10










  • $begingroup$
    Yes, but didn't I integrate from 0 to $2pi$ anyway?
    $endgroup$
    – khaled014z
    Dec 25 '18 at 2:25






  • 1




    $begingroup$
    This seems to be full of errors. Why are you varying $rho$. It's constant, you're on a sphere. The surface element should be $a^2sinthetaoperatorname dthetaoperatorname dvarphi $. You need spherical coordinates, not polar. Etc...
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 3:07














0












0








0





$begingroup$


$iint_S$ r.n $dS$



Over the surface of the sphere with radius $a$ centered at the origin



Now this is obviously trivial and the answer is $4pi a^3$ but I want to do it the hard way because there's something I don't understand



The surface is $x^2 + y^2 + z^2 = a^2$ , then the normal vector $n = nabla S$



$hat n$ = $frac{nabla S}{|nabla S|}$ = $frac{x hat i + y hat j + z hat k}{a}$



$dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$



Then $iint_S$ r.n $dS$ = $iint_S frac{x^2 + y^2}{sqrt{a^2 -x^2 -y^2}} + sqrt{a^2 -y^2 -x^2}$ $dxdy$



Switching to polar coordinates, $x=rho cosphi , y =rho sinphi$



Then $iint_S$ r.n $dS$ = $iint_S frac{rho^2}{sqrt{a^2 -rho^2}} + sqrt{a^2 - rho^2}$ $rho drho dphi$



Integrating $rho$ from $0$ to $a$ and $phi$ from $0$ to $2pi$ , we get:



$iint_S$ r.n $dS$ = $2pi a^3$ which is half the required answer $4pi a^3$ , is it because I only took into account that $dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$ and should have changed this surface element starting from a specific point? If so, how? Thanks






vector-analysis surface-integrals







share|cite|improve this question









$endgroup$




$iint_S$ r.n $dS$



Over the surface of the sphere with radius $a$ centered at the origin



Now this is obviously trivial and the answer is $4pi a^3$ but I want to do it the hard way because there's something I don't understand



The surface is $x^2 + y^2 + z^2 = a^2$ , then the normal vector $n = nabla S$



$hat n$ = $frac{nabla S}{|nabla S|}$ = $frac{x hat i + y hat j + z hat k}{a}$



$dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$



Then $iint_S$ r.n $dS$ = $iint_S frac{x^2 + y^2}{sqrt{a^2 -x^2 -y^2}} + sqrt{a^2 -y^2 -x^2}$ $dxdy$



Switching to polar coordinates, $x=rho cosphi , y =rho sinphi$



Then $iint_S$ r.n $dS$ = $iint_S frac{rho^2}{sqrt{a^2 -rho^2}} + sqrt{a^2 - rho^2}$ $rho drho dphi$



Integrating $rho$ from $0$ to $a$ and $phi$ from $0$ to $2pi$ , we get:



$iint_S$ r.n $dS$ = $2pi a^3$ which is half the required answer $4pi a^3$ , is it because I only took into account that $dS = frac{dxdy}{|hat n . hat k|} = frac{dxdy}{a/z}$ and should have changed this surface element starting from a specific point? If so, how? Thanks






vector-analysis surface-integrals




vector-analysis surface-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 25 '18 at 1:26









khaled014zkhaled014z

1769




1769












  • $begingroup$
    When you take the square root, aren’t you only giving the correct value of $z$ in upper half space?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:10










  • $begingroup$
    Yes, but didn't I integrate from 0 to $2pi$ anyway?
    $endgroup$
    – khaled014z
    Dec 25 '18 at 2:25






  • 1




    $begingroup$
    This seems to be full of errors. Why are you varying $rho$. It's constant, you're on a sphere. The surface element should be $a^2sinthetaoperatorname dthetaoperatorname dvarphi $. You need spherical coordinates, not polar. Etc...
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 3:07


















  • $begingroup$
    When you take the square root, aren’t you only giving the correct value of $z$ in upper half space?
    $endgroup$
    – Charlie Frohman
    Dec 25 '18 at 2:10










  • $begingroup$
    Yes, but didn't I integrate from 0 to $2pi$ anyway?
    $endgroup$
    – khaled014z
    Dec 25 '18 at 2:25






  • 1




    $begingroup$
    This seems to be full of errors. Why are you varying $rho$. It's constant, you're on a sphere. The surface element should be $a^2sinthetaoperatorname dthetaoperatorname dvarphi $. You need spherical coordinates, not polar. Etc...
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 3:07
















$begingroup$
When you take the square root, aren’t you only giving the correct value of $z$ in upper half space?
$endgroup$
– Charlie Frohman
Dec 25 '18 at 2:10




$begingroup$
When you take the square root, aren’t you only giving the correct value of $z$ in upper half space?
$endgroup$
– Charlie Frohman
Dec 25 '18 at 2:10












$begingroup$
Yes, but didn't I integrate from 0 to $2pi$ anyway?
$endgroup$
– khaled014z
Dec 25 '18 at 2:25




$begingroup$
Yes, but didn't I integrate from 0 to $2pi$ anyway?
$endgroup$
– khaled014z
Dec 25 '18 at 2:25




1




1




$begingroup$
This seems to be full of errors. Why are you varying $rho$. It's constant, you're on a sphere. The surface element should be $a^2sinthetaoperatorname dthetaoperatorname dvarphi $. You need spherical coordinates, not polar. Etc...
$endgroup$
– Chris Custer
Dec 25 '18 at 3:07




$begingroup$
This seems to be full of errors. Why are you varying $rho$. It's constant, you're on a sphere. The surface element should be $a^2sinthetaoperatorname dthetaoperatorname dvarphi $. You need spherical coordinates, not polar. Etc...
$endgroup$
– Chris Custer
Dec 25 '18 at 3:07










2 Answers
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$begingroup$

notice that
$$vec r cdot vec n = frac{x^2+y^2+z^2} a = frac {a^2} a = a$$



which is a constant so can be taken outside the integral



so
$$iint_S vec r cdot vec n ;dS = a iint_S ;dS $$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You get $aintintoperatorname dS=aintint a^2sinthetaoperatorname dthetaoperatorname dvarphi=a^3int_0^{2pi}int_0^{pi}sinthetaoperatorname dthetaoperatorname dvarphi=2pi a^3[-costheta]_0^{pi}=4pi a^3$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      0












      $begingroup$

      notice that
      $$vec r cdot vec n = frac{x^2+y^2+z^2} a = frac {a^2} a = a$$



      which is a constant so can be taken outside the integral



      so
      $$iint_S vec r cdot vec n ;dS = a iint_S ;dS $$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        notice that
        $$vec r cdot vec n = frac{x^2+y^2+z^2} a = frac {a^2} a = a$$



        which is a constant so can be taken outside the integral



        so
        $$iint_S vec r cdot vec n ;dS = a iint_S ;dS $$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          notice that
          $$vec r cdot vec n = frac{x^2+y^2+z^2} a = frac {a^2} a = a$$



          which is a constant so can be taken outside the integral



          so
          $$iint_S vec r cdot vec n ;dS = a iint_S ;dS $$






          share|cite|improve this answer









          $endgroup$



          notice that
          $$vec r cdot vec n = frac{x^2+y^2+z^2} a = frac {a^2} a = a$$



          which is a constant so can be taken outside the integral



          so
          $$iint_S vec r cdot vec n ;dS = a iint_S ;dS $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 3:24









          WW1WW1

          7,3251712




          7,3251712























              0












              $begingroup$

              You get $aintintoperatorname dS=aintint a^2sinthetaoperatorname dthetaoperatorname dvarphi=a^3int_0^{2pi}int_0^{pi}sinthetaoperatorname dthetaoperatorname dvarphi=2pi a^3[-costheta]_0^{pi}=4pi a^3$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You get $aintintoperatorname dS=aintint a^2sinthetaoperatorname dthetaoperatorname dvarphi=a^3int_0^{2pi}int_0^{pi}sinthetaoperatorname dthetaoperatorname dvarphi=2pi a^3[-costheta]_0^{pi}=4pi a^3$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You get $aintintoperatorname dS=aintint a^2sinthetaoperatorname dthetaoperatorname dvarphi=a^3int_0^{2pi}int_0^{pi}sinthetaoperatorname dthetaoperatorname dvarphi=2pi a^3[-costheta]_0^{pi}=4pi a^3$.






                  share|cite|improve this answer









                  $endgroup$



                  You get $aintintoperatorname dS=aintint a^2sinthetaoperatorname dthetaoperatorname dvarphi=a^3int_0^{2pi}int_0^{pi}sinthetaoperatorname dthetaoperatorname dvarphi=2pi a^3[-costheta]_0^{pi}=4pi a^3$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 3:39









                  Chris CusterChris Custer

                  14.2k3827




                  14.2k3827






























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